An identical book is dropped on four planets of unknown gravity. The book is thrown down on planets W and X, and released from rest on planets Y and Z. The velocity is recorded for some time intervals on each planet, but not all. The missing velocities are blocked out with a gray box. Which of the following lists the planets in order from the greatest acceleration of gravity to the least acceleration of gravity?
Answers
We want to compare the gravitational acceleration of different planets by using the given table. The correct option is the last one, so we have:
Planet Y, planet W, planet Z, planet X.
First, we know that the velocity equation is written as:
V(t) = a*t + v₀
Where a is the acceleration and v₀ is the initial velocity.
With this equation, we can find the acceleration in each planet.
Planet W
The initial velocity is v₀ = -3m/s
V(t) = a*t - 3m/s
And we know that at t = 2s, the velocity is -22m/s.
v(2s) = -22m/s = a*2s - 3m/s
-22m/s + 3m/s = a*2s
-19 m/s = a*2s
(-19 m/s)/2s = a = -9.5m/s^2
Now we just do the same thing for each planet.
Planet X:
v₀ = -0.5 m/s
v(1s) = -4.2 m/s = a*1s - 0.5 m/s
-4.2 m/s + 0.5 m/s = a*1s
-3.8 m/s^2 = a
Planet Y:
v₀ = 0 m/s
v(2s) = -21 m/s = a*2s
(-21 m/s)/2s = a = -10.5 m/s^2
Planet Z:
v₀ = 0 m/s
v(1s) = -8 m/s = a*1s
(-8 m/s)/1s = a = -8m/s^2
Then the order from greatest to least is:
Y, W, Z, X
So the correct option is the last one.
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Answer:
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Answer:
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Answer:
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Explanation:
The matter through which a mechanical wave travels is called the medium (plural, media). There are three types of mechanical waves: transverse, longitudinal, and surface waves. They differ in how particles of the medium move when the energy of the wave passes through. The particles in a medium become disturbed and pass on this disturbance to the next one, and so on. So, the disturbance caused by the particles can travel through a medium such as air, ocean water, land, violin strings, Slinky coils, etc. Sound waves can travel through solids, liquids, or gases.
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vu
LO
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[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
The Correct choice is ~
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Answer:
1) Work is a measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement. 2) Work is calculated through joules. 3) It's possible to calculate work by multiplying the amount of force applied by the distance that the force is applied. If we know the amount of force the stick applies to the puck and the distance that the stick is applying the force to the puck, we can figure out the amount of work done.
Explanation:
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Plug in the numbers:
Answer (with Units):
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A body of mass 3kg and volume 4×10^-4m^3 is hung from balance graduated in newtons. What is the reading of the balance when the body is;
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We employed the pressure P formulas. the reading of the balance when the body is;
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Given that a body of mass 3kg and volume 4×10^-4m^3 is hung from balance graduated in newtons. To know the reading of the balance when the body is;
a. In air
b. Fully immersed in water
c. ⅓ immersed in water
d. Fully immersed in paraffin.
We will employ the pressure P formulas.
P = F/A
and
P = ρgh
Where
ρ = density
g = acceleration due to gravity
h = height
F = force
A = area
F/A = ρgh
Cross multiply
F = ρghA
Where hA = Volume V
Therefore, F = ρgV
a. In air
F = mg
F = 3 x 9.8
F = 29.4 N
b. Fully immersed in water
F = ρgV
F = [tex]10^{3}[/tex] x 9.8 x 4 x [tex]10^{-4}[/tex]
F = 3.92 N
c. ⅓ immersed in water
F = (ρgV)/3 + 2/3(mg)
F = 3.92/3 + (2/3 x 29.4)
F = 1.3067 + 19.6
F = 20.9N
d. Fully immersed in paraffin.
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F = 800 x 9.8 x 4 x [tex]10^{-4}[/tex]
F = 3.136 N
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Answer:
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mass (m) = 75kg
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Answers
Answer:
735 N
Explanation:
A machine launches a tennis ball at an angle of 45° with the horizontal, as
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0.p2 second after its launch. [Neglect air resistance and assume the ball lands at the
same height above the ground from which it was launched.]
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+0.92 s
Viy = 9.0 m/s
Launcher
V = 9.0 m/s
45
Horizontal
The total horizontal distance traveled by the tennis ball during the entire time it is in the air is
Answers
The motion or path of the tennis ball is the path of a projectile, which has
both vertical and horizontal motion.
The total horizontal distance traveled by the tennis ball during the entire
time it is in the air ≈ 16.514 meters.
Reasons:
The given parameters are;
The direction in which the ball is launched = 45°
The initial vertical velocity of the ball, [tex]v_y[/tex] = 9.0 m/s
The initial horizontal velocity of the ball, vₓ = 9.0 m/s
The time it takes the ball to reach the maximum height, t = 0.92 seconds after its launch
The total horizontal distance traveled, by the tennis ball is given by the formula;
[tex]\Delta d_x = \dfrac{v_1^2 \cdot \left( 2 \cdot cos(\theta) \sin(\theta)}{y} = \dfrac{2 \times v_x \times v_y}{g}[/tex]
Which gives;
[tex]\mathrm{The \ total \ distance \ travelled, \ \Delta} d_x = \dfrac{2 \times 9.0\times 9.0}{9.81} \approx 16.514[/tex]
The total horizontal distance traveled by the tennis ≈ 16.514 meters
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Could someone please help
Answers
Answer: Dispersion
Explanation:
Si olvidas un trozo de chocolate cerca de la estufa encendida, al rato, observa tu chocolate derretido, ¿Cómo estarán las partículas?
Answers
Answer:
partículas estarían separadas
Explanation:
ya que con el calor las partículas se alborotan y con frío se congelan
with what force will Annabella’s car hit a tree if her car has a mass of 5000 kg and an acceleration of 2 m/s2
Answers
Answer:
F=10,000N
Explanation:
F=ma
m=5000kg
a=2m/s^2
[tex]F=(5000kg)*(2m/s^{2} )\\\\\\\\F=10000N[/tex]
Answer:
[tex]\boxed {\boxed {\sf 10,000 \ Newtons}}[/tex]
Explanation:
We are asked to find the force of Annabella's car when it hits a tree.
According to Newton's Second Law of Motion, force is the product of mass and acceleration. Therefore, it is calculated using the following formula:
[tex]F=ma[/tex]
The mass of the car is 5000 kilograms and the acceleration is 2 meters per second squared.
m= 5000 kg a= 2 m/s²Substitute the values into the formula.
[tex]F= (5000 \ kg)(2 \ m/s^2)[/tex]
Multiply.
[tex]F= 10,000 \ kg*m/s^2[/tex]
Convert the units. 1 kilogram meter per second squared is equal to 1 Newton, so our answer of 10,000 kg*m/s² is equal to 10,000 Newtons.
[tex]F= 10,000 \ N[/tex]
Annabella's car hits the tree with a force of 10,000 Newtons.
Jorge is driving along Route 120 at a speed of 25 m/s. Seeing a red light ahead, he applies the brakes and comes to rest in 3.7 seconds. What is
his rate of acceleration?
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Answer:
I don't know...
I'm sorry
A rock is thrown upward with a speed of 2.0m/s.
a. How long will it take the rock to reach the peak?
b. What is the maximum height it can reach?
Answers
Hi there!
(A)
When the rock reaches the peak, it will have a vertical velocity of 0 m/s. We can use the equation:
vf = vi + at
vf = final velocity (0 m/s)
vi = initial velocity (2 m/s)
a = acceleration (in this situation, due to gravity = 9.8 m/s²)
t = time (s)
0 = 2 + (-9.8)t
2/9.8 = t = .204 sec
(B)
Use the equation:
d = vit + 1/2at²
d = 2(.204) + 1/2(-9.8)(.204²)
d = .204 m
Are these pictures below an example of reflection or refraction?
Answers
Answer:
Refraction
Explanation:
Reflection involves a change in direction of waves when they bounce off a barrier.
Refraction of waves involves a change in the direction of waves as they pass from one medium to another.
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2. A student slides a 0.75 kg textbook across a table, and it comes to rest after
traveling 1.2 m. Given that the coefficient of kinetic friction between the
book and the table is 0.34, use the work-kinetic energy theorem to find
the book's initial speed.
Answers
The book's initial speed = 2.83 m/s
The mass of the textbook, m = 0.75 kg
The final speed of the textbook, v = 0 m/s (Since it comes to rest)
The distance covered by the book, d = 1.2 m
Coefficient of kinetic friction, [tex]\mu = 0.34[/tex]
The frictional force between the table and the book is calculated as:
[tex]f_k=-\mu mg\\\\ f_k=-0.34 \times 0.75 \times 9.8\\\\f_k=-2.499N[/tex]
The workdone by the textbook is calculated as:
[tex]W = Fd\\\\W=2.499(1.2)\\\\W=2.9988J[/tex]
According to the work-kinetic energy theorem:
Workdone by the book = Change in Kinetic energy
[tex]W = 0.5m(v^2-u^2)\\\\-2.9988=0.5 \times 0.75 \times (0^2-u^2)\\\\-2.99988=-0.375u^2\\\\u^2=\frac{-2.9988}{-0.375} \\\\u^2=7.9968\\\\u=\sqrt{7.9968}\\\\u=2.83m/s[/tex]
The book's initial speed = 2.83 m/s
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